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UPSC CDS II 2025 Mathematics Paper

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Question : 40 of 100

Marks: +1, -0

What is

\frac{\frac{x}{x-y}+\frac{y}{y-z}+\frac{z}{z-x}}{\frac{x+y}{x-y}+\frac{y+z}{y-z}+\frac{z+x}{z-x}+3}

equal to?

1
$1/2$
$1/3$
$1/4$

Solution:

Concept:

Rewrite each fraction using the identity

\frac{y}{x-y} = \frac{x}{x-y} - 1

to relate numerator and denominator.

Explanation:

Let

A = \frac{x}{x-y} + \frac{y}{y-z} + \frac{z}{z-x}

.

Consider the sum

\frac{y}{x-y} + \frac{z}{y-z} + \frac{x}{z-x}

.

Using

\frac{y}{x-y} = \frac{x}{x-y} - 1

, similarly for other terms:

This sum becomes

\left(\frac{x}{x-y} - 1\right) + \left(\frac{y}{y-z} - 1\right) + \left(\frac{z}{z-x} - 1\right) = A - 3

.

Now the denominator

D = \frac{x+y}{x-y} + \frac{y+z}{y-z} + \frac{z+x}{z-x} + 3

.

Write each term as sum:

\frac{x+y}{x-y} = \frac{x}{x-y} + \frac{y}{x-y}

.

So

D = \left(\frac{x}{x-y} + \frac{y}{y-z} + \frac{z}{z-x}\right) + \left(\frac{y}{x-y} + \frac{z}{y-z} + \frac{x}{z-x}\right) + 3

.

Substitute:

D = A + (A - 3) + 3 = 2A

.

Thus the given expression

= \frac{A}{2A} = \frac{1}{2}

.

Answer:

Option B:

\frac{1}{2}

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